Problem: $\lim_{x\to\infty}\dfrac{7x^2}{x-x^3}=?$ Choose 1 answer: Choose 1 answer: (Choice A) A $-7$ (Choice B) B $0$ (Choice C) C $7$ (Choice D) D $-\infty$
$\lim_{x\to\infty} 7x^2=\infty$ and $\lim_{x\to\infty} x-x^3=-\infty$, so $\lim_{x\to\infty}\dfrac{7x^2}{x-x^3}$ results in the indeterminate form $\dfrac{\infty}{-\infty}$. We should use l'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to\infty}\dfrac{7x^2}{x-x^3} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[7x^2\right]}{\dfrac{d}{dx}[x-x^3]} \gray{\text{l'Hôpital's rule}} \\\\ &=\lim_{x\to\infty}\dfrac{14x}{1-3x^2} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[14x\right]}{\dfrac{d}{dx}[1-3x^2]} \gray{\text{l'Hôpital's rule}} \\\\ &=\lim_{x\to\infty}\dfrac{14}{-6x} \\\\ &=0 \end{aligned}$ Note that we used l'Hôpital's rule twice, because the first time we used it, we ended with the indeterminate form $\dfrac{\infty}{-\infty}$ too. Also note that we were only able to use l'Hôpital's rule because the limit $\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[14x\right]}{\dfrac{d}{dx}[1-3x^2]}$ can actually be determined. In conclusion, $\lim_{x\to\infty}\dfrac{7x^2}{x-x^3}=0$.